Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[5]{{5x + 1}} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {5x + 1 - 1} \right)}}{{x.\left( {{{\left( {\sqrt[5]{{5x + 1}}} \right)}^4} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^3} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^2} + \sqrt[5]{{5x + 1}} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{{{\left( {\sqrt[5]{{5x + 1}}} \right)}^4} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^3} + {{\left( {\sqrt[5]{{5x + 1}}} \right)}^2} + \sqrt[5]{{5x + 1}} + 1}}\\
= \dfrac{1}{{1 + 1 + 1 + 1 + 1}}\\
= \dfrac{1}{4}\\
\mathop {\lim }\limits_{x \to 8} \dfrac{{\sqrt {9 + 2x} - 5}}{{\sqrt[3]{x} - 2}}\\
= \mathop {\lim }\limits_{x \to 8} \dfrac{{\left( {9 + 2x - 25} \right)\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\left( {x - 8} \right)\left( {\sqrt {9 + 2x} + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 8} \dfrac{{2.\left( {x - 8} \right)\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\left( {x - 8} \right)\left( {\sqrt {9 + 2x} + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 8} \dfrac{{2\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\sqrt {9 + 2x} + 5}}\\
= \dfrac{{2.\left( {\sqrt[3]{{{8^2}}} + 2.\sqrt[3]{8} + 4} \right)}}{{\sqrt {9 + 2.8} + 5}}\\
= \dfrac{{2.\left( {4 + 4 + 4} \right)}}{{5 + 5}}\\
= \dfrac{{12}}{5}
\end{array}$
